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Gay Lussac's Law, Avagadro's Law

19 grams of a...

Question

$19$ grams of a mixture containing $N a H C O_{3}$ and $N a_{2} C O_{3}$ on complete heating liberated $1.12 l i t$ of $C O_{2}$ at STP. The weight of the remaining solid was $15.9 g$ . What is the weight (in g) of $N a_{2} C O_{3}$ in the mixture before heating?

8.4

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15.9

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4.0

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10.6

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Solution

The correct option is C $10.6$
$2 N a H C O_{3} Δ - \to N a_{2} C O_{3} + H_{2} O + C O_{2}$
1.12 L $C O_{2}$ at STP corresponds to $\frac{1.12 L}{22.4 L / m o l} = 0.05$ moles.
0.05 moles of $C O_{2}$ will come from $2 \times 0.05 = 0.1$ mole of $N a H C O_{3}$ .
The molar mass of $N a H C O_{3}$ is 84 g/mol. Mass of 0.1 moles of $N a H C O_{3} = 84 \times 0.1 = 8.4 g$
Mass of $N a_{2} C O_{3} = 19 - 8.4 = 10.6 g$

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